Limits at infinity and horizontal asymptotes

Teuta proves a limit from the definition

Teuta wants to prove that \lim_{x \to \infty} \dfrac{4x + 1}{x + 3} = 4 directly from the definition. Find an N for the tolerance t = 0.01, and then one that works for every t > 0.

The definition asks about the distance |f(x) - 4|, so compute it. For x > -3,

so \displaystyle |f(x) - 4| = \frac{11}{x + 3}, since x + 3 > 0. This is less than t exactly when \displaystyle x + 3 > \frac{11}{t}, that is, when \displaystyle x > \frac{11}{t} - 3.

For t = 0.01 this gives x > 1100 - 3 = 1097, so N = 1097 works. For a general tolerance, \displaystyle N = \frac{11}{t} - 3 works, since every x > N has |f(x) - 4| < t. An N exists for every t > 0, which is what the definition demands, so the limit is 4. Figure 1 shows the case t = 1, where N = 11 - 3 = 8.

The sign of \displaystyle \frac{-11}{x + 3} says more: for x > -3 the outputs are always below 4, so the graph approaches its asymptote y = 4 from below.

Dividing by the highest power

Find \lim_{x \to \infty} \dfrac{4x^2 - 3x}{2x^2 + 5}, and the limit as x \to -\infty.

The quotient law cannot be used as the function stands: top and bottom both grow without bound, so neither has a limit that is a number. Divide every term of the top and the bottom by x^2, the highest power in the bottom. This is allowed because x \ne 0 for large x, and dividing top and bottom by the same nonzero number does not change a fraction:

By the basic limit, \displaystyle \frac{3}{x} \to 0 and \displaystyle \frac{5}{x^2} \to 0 (constant multiple law). The difference and sum laws give top \to 4 and bottom \to 2, and since 2 \ne 0 the quotient law gives

The powers are positive integers, so \displaystyle \frac{3}{x} and \displaystyle \frac{5}{x^2} also go to 0 as x \to -\infty, and the limit there is 2 as well. The graph has the single horizontal asymptote y = 2, approached at both ends. A check: at x = 100 the function is \displaystyle \frac{39{,}700}{20{,}005} \approx 1.9845.

Dren derives the degree comparison

Dren knows the rule in Horizontal asymptotes for reading the asymptote of a rational function from the degrees of its top and bottom. Derive it from the limit laws for

Divide every term of top and bottom by x^m. The bottom becomes \displaystyle b_m + \frac{b_{m-1}}{x} + \cdots + \frac{b_0}{x^m}, which tends to b_m \ne 0 as x \to \infty or x \to -\infty, by the basic limit with positive integer powers. Each top term becomes a_k x^{k - m}.

• Top degree smaller, n < m. Every k \le n < m, so each top term is \displaystyle \frac{a_k}{x^{m-k}} \to 0. The limit is \displaystyle \frac{0}{b_m} = 0, and y = 0 is the asymptote.
• Equal degrees, n = m. The term a_n x^0 = a_n stays and every other term goes to 0, so the limit is \displaystyle \frac{a_n}{b_m}, the ratio of leading coefficients.
• Top degree larger, n > m. Divide by x^n instead. Now the top tends to a_n, while every bottom term has a positive power of x below it, so the bottom tends to 0. A top near a_n \ne 0 over a bottom near 0 grows without bound in size, so there is no limit that is a number and no horizontal asymptote.

The same three answers hold as x \to -\infty, so a rational function with a horizontal asymptote has the same one at both ends. Example 5 shows a function that is not rational where the two ends differ.

A fractional power that settles slowly

Find \displaystyle \lim_{x \to \infty} \dfrac{2x - \sqrt{x}}{3x + 4\sqrt{x}}.

Write \displaystyle \sqrt{x} = x^{1/2}. The highest power present is x^1, so divide every term by x, using \displaystyle \frac{x^{1/2}}{x} = \frac{1}{x^{1/2}}:

The basic limit holds for \displaystyle p = \frac{1}{2}, so \displaystyle \frac{1}{\sqrt{x}} \to 0, and the laws give

The limit as x \to -\infty is not asked, and cannot be: \displaystyle \sqrt{x} is undefined for x < 0. The approach is slow. At x = 10{,}000, where \displaystyle \sqrt{x} = 100, the value is \displaystyle \frac{19{,}900}{30{,}400} \approx 0.6546, still more than 0.01 from \displaystyle \frac{2}{3} \approx 0.6667, because \displaystyle \frac{1}{\sqrt{x}} shrinks much more slowly than \displaystyle \frac{1}{x}. A table could suggest a wrong value here; the algebra settles it.

One function, two horizontal asymptotes

Find \displaystyle \lim_{x \to \infty} \dfrac{\sqrt{9x^2 + 1}}{x - 2} and \displaystyle \lim_{x \to -\infty} \dfrac{\sqrt{9x^2 + 1}}{x - 2}.

To divide by x, the x has to come out of the square root, and that is where the sign enters. For x \ne 0, factor x^2 out under the root and use \displaystyle \sqrt{x^2} = |x| (the move of Simplifying radicals):

As x \to \infty, x > 0 and |x| = x. Cancel the x against the x factored from the bottom, \displaystyle x - 2 = x\left(1 - \frac{2}{x}\right):

The root law applies because \displaystyle 9 + \frac{1}{x^2} \to 9 > 0.

As x \to -\infty, x < 0 and |x| = -x, so a minus sign survives the cancellation:

The graph has two horizontal asymptotes, y = 3 on the right and y = -3 on the left (Figure 2). This makes sense: the top is always positive, while the bottom is positive on the far right and negative on the far left. The line x = 2, where the bottom is 0 and the top is not, is a vertical asymptote, treated in Infinite limits and vertical asymptotes.

An oblique asymptote from two limits

Find the oblique asymptote of f(x) = \dfrac{2x^2 + 3x - 1}{x + 2} by limits, and check it by division.

First find m by dividing top and bottom by x^2:

Then find b by putting f(x) - 2x over the common bottom:

So b = -1 and the oblique asymptote is y = 2x - 1. Every limit here is the same as x \to -\infty, so the same line serves at both ends.

As a check, Polynomial long division divides 2x^2 + 3x - 1 by x + 2 to give quotient 2x - 1 and remainder 1, since (x + 2)(2x - 1) = 2x^2 + 3x - 2. So \displaystyle f(x) - (2x - 1) = \frac{1}{x + 2}, which goes to 0, is positive for x > -2 and negative for x < -2: the graph lies above the line on the right and below it on the left. Division works only for rational functions; the two limits work for any function, as the next example shows. The division method itself is Slant asymptotes.

Vlora’s curve with two oblique asymptotes

Vlora graphs \displaystyle h(x) = \sqrt{x^2 + 4x}, which is defined for x \le -4 and x \ge 0. Find its oblique asymptotes, including \displaystyle \lim_{x \to \infty} \big(\sqrt{x^2 + 4x} - x\big).

The right end. For x > 0, \displaystyle \sqrt{x^2 + 4x} = x\sqrt{1 + \frac{4}{x}}, so

Next, \displaystyle b = \lim_{x \to \infty} \big(\sqrt{x^2 + 4x} - x\big). Both pieces grow without bound, and the difference law needs numbers, so rewrite first. Multiply top and bottom by the conjugate \displaystyle \sqrt{x^2 + 4x} + x, the move from Limits by algebra: factoring and rationalizing; the difference of squares clears the root:

So b = 2, and y = x + 2 is the asymptote on the right.

The left end. For x < 0, \displaystyle \sqrt{x^2 + 4x} = |x|\sqrt{1 + \frac{4}{x}} = -x\sqrt{1 + \frac{4}{x}}, so \displaystyle m = \lim_{x \to -\infty} \left(-\sqrt{1 + \frac{4}{x}}\right) = -1. Then \displaystyle b = \lim_{x \to -\infty} \big(\sqrt{x^2 + 4x} + x\big), and the conjugate is \displaystyle \sqrt{x^2 + 4x} - x:

So y = -x - 2 is the asymptote on the left (Figure 3). Note that \displaystyle \sqrt{x^2 + 4x} - x itself has no limit as x \to -\infty: both terms are positive and large, so their sum grows without bound, and no conjugate is needed to see it.

The two lines cross at (-2, 0), and the graph is symmetric about x = -2, since x^2 + 4x = (x + 2)^2 - 4. That identity also shows h(x) < |x + 2| wherever h is defined, so the graph runs just under both lines.